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| Date | Wed, 18 Feb 2015 15:54:08 +1100 |
| Subject | Re: Noob Parsing question |
| From | Chris Angelico <rosuav@gmail.com> |
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Show key headers only | View raw
On Wed, Feb 18, 2015 at 3:35 PM, <kai.peters@gmail.com> wrote:
>> > Given
>> >
>> > data = '{[<a=14^b=Fred^c=45.22^><a=22^b=Joe^><a=17^c=3.20^>][<a=72^b=Soup^>]}'
>> >
>> > How can I efficiently get dictionaries for each of the data blocks framed by <> ?
>> >
>> > Thanks for any help
>>
>> The question here is: What _can't_ happen? For instance, what happens
>> if Fred's name contains a greater-than symbol, or a caret?
>>
>> If those absolutely cannot happen, your parser can be fairly
>> straight-forward. Just put together some basic splitting (maybe a
>> regex), and then split on the caret inside that. Otherwise, you may
>> need a more stateful parser.
>>
>> ChrisA
>
> The data string is guaranteed to be clean - no such irregularities occur.
Okay!
(Side point: You've stripped off all citations, here, so it's not
clear who said what. My shorthand signature isn't as useful as the
full line identifying date, time, and person. It's polite to keep
those lines, at least for the first level of quoting.)
What you want can be done with a regular expression. (Yes, yes, I
know; now you have two problems.)
>>> data = '{[<a=14^b=Fred^c=45.22^><a=22^b=Joe^><a=17^c=3.20^>][<a=72^b=Soup^>]}'
>>> re.findall("<.*?>",data)
['<a=14^b=Fred^c=45.22^>', '<a=22^b=Joe^>', '<a=17^c=3.20^>', '<a=72^b=Soup^>']
>From there, you can crack open the different pieces:
>>> for piece in re.findall("<.*?>",data):
... d = {}
... for elem in piece[1:-2].split("^"):
... key, value = elem.split("=",1)
... d[key] = value
... print(d)
...
{'c': '45.22', 'b': 'Fred', 'a': '14'}
{'b': 'Joe', 'a': '22'}
{'c': '3.20', 'a': '17'}
{'b': 'Soup', 'a': '72'}
If you need some of those to be integers or floats, you'll need to do
some post-processing on it, but this guarantees that you get the data
out reliably. It depends on not having any of the special characters
"=^<>" inside the elements, but other than that, it should be safe.
ChrisA
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Noob Parsing question kai.peters@gmail.com - 2015-02-17 20:07 -0800
Re: Noob Parsing question Chris Angelico <rosuav@gmail.com> - 2015-02-18 15:16 +1100
Re: Noob Parsing question kai.peters@gmail.com - 2015-02-17 20:35 -0800
Re: Noob Parsing question Chris Angelico <rosuav@gmail.com> - 2015-02-18 15:54 +1100
Re: Noob Parsing question kai.peters@gmail.com - 2015-02-18 08:57 -0800
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